A fox one day invited a stork to dinner, and being disposed to divert himself at the expense of his — Analytical Chemistry Chemistry Question
The fox and the stork
A fox one day invited a stork to dinner, and being disposed to divert himself at the expense of his guest, provided nothing for the entertainment but some thin soup in a shallow dish. This the fox lapped up very readily, while the stork, unable to gain a mouthful with her long narrow bill, was as hungry at the end of dinner as when she began. The fox meanwhile professed his regret at seeing her eat so sparingly and feared that the dish was not seasoned to her mind. [VISUAL]
The stork said little, but begged that the fox would do her the honour of repaying her visit. Accordingly, he agreed to dine with her on the following day. He arrived true to his appointment and the dinner was ordered forthwith.
When the meal was served up, the fox found to his dismay that it was contained in a narrow-necked vessel, down which the stork readily thrust her long neck and bill, while he was obliged to content himself with licking the neck of the jar. Unable to satisfy his hunger, he retired with as good a grace as he could, observing that he could hardly find fault with his entertainer, who had only paid him back in his own coin. [VISUAL]
An alternative end of the fable (instead of the grey sentence) could be:
However, as the fox was very clever, he took a look around and found a solution to his problem. There were many pebbles lying around. The fox did not hesitate even for a moment and started to throw them inside the jar of soup. The stork was shaking her head in confusion as the fox kept throwing the pebbles in, until the moment that the surface of the soup reached the brim of the jar. Then the fox turned to the stork with a smirk on his face and said: “Of course I will taste it,” and started to eat the soup.
A minimum volume of the soup in the jar is necessary for the fox to succeed. This volume is related to the total volume of the pebbles eventually present in the jar. This total volume is related to the number, size and way of arrangement of the pebbles.
Let us approximate the situation by a geometrical model:
- The jar is approximated as a perfect cylinder with a diameter of 10.0 cm and a height of 50.0 cm.
- A pebble is approximated as a perfect hard-sphere.
- All the spheres have the same diameter.
- The spheres are arranged as close as possible so that they touch each other.
- The soup is approximated by water.
- All pebbles are fully inside the jar (i.e. no part of any pebble is above the rim of the cylinder).
Calculate the maximum number of spheres that fit into the cylinder. (Under section 'Large stones', where the radius of the sphere is r = 5 cm.)
Model Answer
Each layer consists of one sphere only: n = 50 / (2 × 5) = 5
Calculate the fraction (in %) of the cylinder volume occupied by this number of spheres. (Under section 'Large stones', where the radius of the sphere is r = 5 cm.)
Model Answer
The volume of 5 spheres: V = 5 × 4/3 πr^3 = 5 × 4/3 π × 5^3 = 2 618 cm^3
The volume of cylinder: V = πr^2 × v = π × 5^2 × 50 = 3 927 cm^3
The fraction of volume: f = 2 618 / 3 927 = 0.667, i.e. 66.7 %
Calculate the free volume (in cm3) among the spheres that can be filled with water. (Under section 'Large stones', where the radius of the sphere is r = 5 cm.)
Model Answer
The free volume: Vfree = 3 927 − 2 618 = 1 309 cm^3
Calculate the radius of the sphere (in cm) considering an arrangement in which 7 spheres in the first (base) layer just fit into the cylinder: [VISUAL]
Model Answer
Radius: r = 10 / 6 = 1.667 cm
Calculate the maximum number of layers that fit in the cylinder (considering an arrangement in which all the higher layers copy the positions of the spheres in the base layer).
Model Answer
Number of layers: N = 50 / (2 × 1.667) = 15
Calculate the maximum number of spheres that fit in the cylinder (considering an arrangement in which all the higher layers copy the positions of the spheres in the base layer).
Model Answer
Number of spheres: n = 15 × 7 = 105
Calculate the fraction (in %) of the cylinder volume occupied by this number of spheres (considering an arrangement in which all the higher layers copy the positions of the spheres in the base layer).
Model Answer
The volume of 105 spheres: V = 105 × 4/3 πr^3 = 105 × 4/3 π × (1.667)^3 = 2 036 cm^3
The fraction of volume: f = 2 036 / 3 927 = 0.518, i.e. 51.8 %
Calculate the free volume (in cm3) among the spheres that can be filled with water (considering an arrangement in which all the higher layers copy the positions of the spheres in the base layer).
Model Answer
The free volume: Vfree = 3 927 − 2 036 = 1 891 cm^3
Calculate the maximum number of layers that fit in the cylinder (considering an arrangement in which each even layer consists of 3 spheres and each odd layer copies the positions of the spheres in the base layer).
Model Answer
The interlayer distance can be calculated as a height of a regular tetrahedron formed by 4 spheres with edge a = 2 × r:
[VISUAL] h = a * sqrt(2/3) = 2 × 1.667 × sqrt(2/3) = 2.722 cm
The distance of the first and the last layer from the bases of the cylinder will be at minimum equal to r. Thus the maximum number of layers:
N = (50 − 2 × r) / h + 1 = (50 − 2 × 1.667) / 2.722 + 1 = 18.14 → 18 layers
Calculate the maximum number of spheres that fit in the cylinder (considering an arrangement in which each even layer consists of 3 spheres and each odd layer copies the positions of the spheres in the base layer).
Model Answer
The total number of spheres: each of the 9 odd layers contains 7 spheres, each of the 9 even layers contains 3 spheres, the total number is:
n = 9 × 7 + 9 × 3 = 90
Calculate the fraction (in %) of the cylinder volume occupied by this number of spheres (considering an arrangement in which each even layer consists of 3 spheres and each odd layer copies the positions of the spheres in the base layer).
Model Answer
The volume of 90 spheres: V = 90 × 4/3 πr^3 = 90 × 4/3 π × (1.667)^3 = 1 745 cm^3
The fraction of volume: f = 1 745 / 3 927 = 0.444, i.e. 44.4 %
Calculate the free volume (in cm3) among the spheres that can be filled with water (considering an arrangement in which each even layer consists of 3 spheres and each odd layer copies the positions of the spheres in the base layer).
Model Answer
The free volume: Vfree = 3 927 − 1 745 = 2 182 cm^3
Calculate the limiting fraction (in %) of the cylinder volume occupied by the spheres, considering very small spheres with diameter smaller by orders of magnitude than the diameter of cylinder (r → 0).
Model Answer
The situation corresponds to the theoretical maximum possible space filling by spheres known as "close-packing of equal spheres". The limiting fraction is:
[VISUAL] f = pi / (3 * sqrt(2)) = 0.7405, i.e. 74.05 %.
There are many ways to derive this ratio. The derivation from a face centered cubic (fcc) elementary cell is shown.
The lattice constant a is 2 × r × √2. Then the volume of the elementary cell is:
Vcell = a^3 = 16 √2 × r^3
The number of spheres belonging to the elementary cell is:
n = 8 × 1/8 (spheres in vertices) + 6 × 1/2 (spheres in the centers of faces) = 4
Thus the fraction volume occupied by the spheres is:
[VISUAL] f = (4 × 4/3 πr^3) / (16 √2 × r^3) = π / (3√2) = 0.7405, i.e. 74.05 %
Calculate the free volume among the spheres that can be filled with water, considering very small spheres with diameter smaller by orders of magnitude than the diameter of cylinder (r → 0).
Model Answer
The free volume: Vfree = 3 927 × (1 − 0.7405) = 1 019 cm^3